Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5r - 10}{r^2 - 5r - 14} $
Solution: First factor the quadratic. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5r - 10}{(r - 7)(r + 2)} $ Then factor out any other terms. $q = \dfrac{r - 7}{r - 3} \times \dfrac{-5(r + 2)}{(r - 7)(r + 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 7) \times -5(r + 2) } { (r - 3) \times (r - 7)(r + 2) } $ $q = \dfrac{ -5(r - 7)(r + 2)}{ (r - 3)(r - 7)(r + 2)} $ Notice that $(r + 2)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -5\cancel{(r - 7)}(r + 2)}{ (r - 3)\cancel{(r - 7)}(r + 2)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $q = \dfrac{ -5\cancel{(r - 7)}\cancel{(r + 2)}}{ (r - 3)\cancel{(r - 7)}\cancel{(r + 2)}} $ We are dividing by $r + 2$ , so $r + 2 \neq 0$ Therefore, $r \neq -2$ $q = \dfrac{-5}{r - 3} ; \space r \neq 7 ; \space r \neq -2 $